Solve x² + 11x - 10 = 0
The solutions of x² + 11x - 10 = 0 are x = (-11 + √161)/2 or x = (-11 - √161)/2, that is x ≈ 0.84428877 and x ≈ -11.844289.
Change a, b or c (whole numbers, a not 0) and the exact solutions follow.
How to work it out
- 1
Write it in standard form a x² + b x + c = 0 and read off the coefficients: a = 1, b = 11, c = -10.
- 2
Work out the discriminant D = b² − 4ac. Its sign tells you how many real solutions there are.
- 3
D is positive, so there are two different real solutions.
- 4
The quadratic formula works for every quadratic:
- 5
Simplify (pull the square out of the root) to get the solutions:
- 6
As decimals: x ≈ 0.84428877 and x ≈ -11.844289.
- 7
Check numerically: for x ≈ 0.84428877 the left side is -1.7763568e-15 (zero up to rounding) ✓
LaTeX source of the formulas on this page
\[ x^2 + 11x - 10 = 0\ \Rightarrow\ x = \frac{-11 + \sqrt{161}}{2},\ \frac{-11 - \sqrt{161}}{2} \]
\[ x^2 + 11x - 10 = 0 \]
\[ D = 11^2 - 4\cdot1\cdot(-10) = 121 + 40 = 161 \]
\[ x = \frac{-b \pm \sqrt{D}}{2a} = \frac{-11 \pm \sqrt{161}}{2} \]
\[ x = \frac{-11 + \sqrt{161}}{2}\quad\text{or}\quad x = \frac{-11 - \sqrt{161}}{2} \]The parabola y = x² + 11x - 10
| x | y |
|---|---|
| -8.5 | -31.25 |
| -7.5 | -36.25 |
| -6.5 | -39.25 |
| -5.5 | -40.25 |
| -4.5 | -39.25 |
| -3.5 | -36.25 |
| -2.5 | -31.25 |
Questions
What are the solutions of x² + 11x - 10 = 0?
The solutions of x² + 11x - 10 = 0 are x = (-11 + √161)/2 or x = (-11 - √161)/2, that is x ≈ 0.84428877 and x ≈ -11.844289.
Does x² + 11x - 10 = 0 factor?
Not over the integers, because the discriminant 161 is not a perfect square. Use the quadratic formula.
What is the vertex of y = x² + 11x - 10?
(-11/2, -161/4). The parabola opens upwards, so this is the minimum.
What are the sum and product of the roots?
Sum = −b/a = -11, product = c/a = -10.